# Is svd of a Gaussian iid matrix corresponds to Haar measure on the Stiefel manifold?

I want to draw a matrix $$A\in \mathbb{R}^{n\times k}$$ uniformally at random from the Stiefel manifold $$\mathbb{V}_k(\mathbb{R}^n)$$, that is from the collection of all $$n\times k$$ matrices $$A$$ such that $$A^TA=I_{k\times k}$$. Is this true that generating a matrix $$X\in\mathbb{R}^{n\times n}$$ with iid standard normal entries $$\mathcal{N}(0,1)$$, performing a svd decomosition $$X=U\Lambda V^H$$ and taking the first $$k$$ vectors of $$U$$, correspond to sampling $$\mathbb{V}_k(\mathbb{R}^n)$$ uniformally at random?

Yes, this is correct; the probability distribution of $$X=U\Lambda V^\top$$ is $$P(X)\propto \exp\left(-\tfrac{1}{2}\,{\rm tr}\,XX^\top\right)=\exp\left(-\tfrac{1}{2}\,{\rm tr}\,\Lambda^2\right),$$ so it is independent of the orthogonal matrices $$U$$,$$V$$. These are therefore distributed uniformly in $$O(n)$$, and identifying $$A$$ with the first $$k$$ columns of $$U$$ will generate a uniformly distributed $$A$$ in $$\mathbb{V}_k(\mathbb{R}^n)$$.
An alternative approach, which does not require you to perform a SVD, is to orthonormalize the first $$k$$ columns of $$X$$ and place these in $$A$$.
• indeed, this non-uniqueness of the SVD can be remedied by multiplying $U$ from the right with a diagonal matrix of random phase factors $e^{i\phi_1},e^{i\phi_2},\ldots e^{i\phi_n}$. Jan 22 at 21:32